` If y =sqrt (tan sqrt x ), then (dy)/(dx) =`

Options

A. `(sec^(2)sqrtx)/( 4sqrt(xtan sqrt x)) `
B. ` (sec ^(2)sqrt x)/( 2sqrt(xtan sqrtx))`
C. ` (sec ^(2)sqrt(x))/(4sqrt(tan x sqrt(x)))`
D. ` (sec ^(2)sqrtx)/(2sqrttan xsqrtx)`

Answer

To find the derivative of \( y = \sqrt{\tan(\sqrt{x})} \), we will use the chain rule and the differentiation formulas for square roots and trigonometric functions. Here is the step-by-step solution:

### Step 1: Identify the outer and inner functions
Let \( u = \tan(\sqrt{x}) \). Then, we can rewrite \( y \) as:
\[ y = \sqrt{u} \]

### Step 2: Differentiate the outer function
Using the formula for the derivative of \( \sqrt{u} \):
\[ \frac{dy}{du} = \frac{1}{2\sqrt{u}} \]

### Step 3: Differentiate the inner function
Now we need to differentiate \( u = \tan(\sqrt{x}) \). Using the chain rule:
\[ \frac{du}{dx} = \sec^2(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) \]
The derivative of \( \sqrt{x} \) is:
\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \]
Thus, we have:
\[ \frac{du}{dx} = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]

### Step 4: Combine the derivatives using the chain rule
Now we can find \( \frac{dy}{dx} \) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\tan(\sqrt{x})}} \cdot \left( \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \right) \]

### Step 5: Simplify the expression
Now we simplify:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\tan(\sqrt{x})}} \cdot \frac{\sec^2(\sqrt{x})}{2\sqrt{x}} = \frac{\sec^2(\sqrt{x})}{4\sqrt{x}\sqrt{\tan(\sqrt{x})}} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{\sec^2(\sqrt{x})}{4\sqrt{x}\sqrt{\tan(\sqrt{x})}} \]